Planar pendulum with rotating pivot

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

$x = x_p + x_m = R\cos(\omega t) + l\sin(\phi)$

$y = -y_p - y_m = -R\sin(\omega t) -l\cos\phi$

For this problem,

My working for finding the coordinates of the mass is,
$x = x_p + x_m = R\cos(\omega t) + l\sin(\phi)$
$y = -y_p - y_m = -R\sin(\omega t) -l\cos\phi$

However, I am told that correct coordinates of the mass is
$x = x_p + x_m = R\cos(\omega t) + l\sin(\phi)$
$y = y_p - y_m = R\sin(\omega t) -l\cos\phi$

I am confused why the y_p is positive since it is clearly negative in the diagram. Can someone please explain to me?

Also if anybody has any tips for finding coordinates in general for physics that would be greatly appreciated! It seems to be the hardest part in mechanics problems.

Thanks!

 

[haruspex]

I am confused why the y_p is positive since it is clearly negative in the diagram. Can someone please explain to me?

The diagram clearly shows that the positive y axis points upwards. In the position shown, $\omega t$ is in the fourth quadrant so $y_p$ will have a negative value.
What does look wrong is the $-y_m$. If $y_m$ is the offset from P to m in the positive y-axis direction then $y=y_p+y_m=R\sin(\omega t)-l\cos(\phi)$.

 

[ChiralSuperfields]

The diagram clearly shows that the positive y axis points upwards. In the position shown, $\omega t$ is in the fourth quadrant so $y_p$ will have a negative value.
What does look wrong is the $-y_m$. If $y_m$ is the offset from P to m in the positive y-axis direction then $y=y_p+y_m=R\sin(\omega t)-l\cos(\phi)$.

Thank you for your reply @haruspex! I agree that $-y_m$ is wrong.

Sorry, are you saying I'm correct and the answer is wrong? i.e $y=-R\sin(\omega t)-l\cos(\phi)$ is indeed correct since you said that $y_p < 0$? If so I should raise with marker.

Thanks!

 

[haruspex]

Sorry, are you saying I'm correct and the answer is wrong?

No. You are confusing the sign of the variable in the equation with the sign of its value in a particular state.
$y, y_p, y_m$ are all measured with up positive. Since $y_p$ is defined as the height of P above the origin and $y_m$ is defined as the height of m above P, it is necessarily true that $y$, the height of m above the origin, is $y_p+y_m$.
In the diagram, $\omega t$ is in the fourth quadrant so its sine is negative and $y_p$ will have a negative value. The way $\phi$ is defined, $y_m=-l\cos(\phi)$, and in the diagrammed position that will also be negative. Adding those two negative values produces a negative value of greater magnitude, exactly as shown in the diagram.

 

[ChiralSuperfields]

No. You are confusing the sign of the variable in the equation with the sign of its value in a particular state.
$y, y_p, y_m$ are all measured with up positive. Since $y_p$ is defined as the height of P above the origin and $y_m$ is defined as the height of m above P, it is necessarily true that $y$, the height of m above the origin, is $y_p+y_m$.
In the diagram, $\omega t$ is in the fourth quadrant so its sine is negative and $y_p$ will have a negative value. The way $\phi$ is defined, $y_m=-l\cos(\phi)$, and in the diagrammed position that will also be negative. Adding those two negative values produces a negative value of greater magnitude, exactly as shown in the diagram.

Thank you for your reply @haruspex!

Sorry I am still confused. The y-coordinate in the first quadrant (not shown in the diagram) is $y=R\sin(\omega t) - l\cos(\phi)$ and in the fourth quadrant (shown in the diagram) is $y=-R\sin(\omega t)-l\cos(\phi)$? Is that please correct?

If each of those coordinates are correct for their respective quadrants, do you please know which are we meant to use for finding the Lagrangian and why?

Thanks!

 

[haruspex]

Thank you for your reply @haruspex!

Sorry I am still confused. The y-coordinate in the first quadrant (not shown in the diagram) is $y=R\sin(\omega t) - l\cos(\phi)$ and in the fourth quadrant (shown in the diagram) is $y=-R\sin(\omega t)-l\cos(\phi)$? Is that please correct?

If each of those coordinates are correct for their respective quadrants, do you please know which are we meant to use for finding the Lagrangian and why?

Thanks!

No, the equation can't magically change. It has to be the one equation for all configurations that arise in the motion.
Consider two positions, first quadrant, $\omega t=\pi/6$ and fourth quadrant, $\omega t=-\pi/6$.
In the first quadrant, $y=R\sin(\omega t) - l\cos(\phi)=R/2- l\cos(\phi)$. In the fourth quadrant, $y=R\sin(\omega t) - l\cos(\phi)=-R/2- l\cos(\phi)$. Same equation, different values.

 

[ChiralSuperfields]

No, the equation can't magically change. It has to be the one equation for all configurations that arise in the motion.
Consider two positions, first quadrant, $\omega t=\pi/6$ and fourth quadrant, $\omega t=-\pi/6$.
In the first quadrant, $y=R\sin(\omega t) - l\cos(\phi)=R/2- l\cos(\phi)$. In the fourth quadrant, $y=R\sin(\omega t) - l\cos(\phi)=-R/2- l\cos(\phi)$. Same equation, different values.

Ooooh thank you @haruspex!

I think I see what you are saying now. Am I please correct to say that we are not meant to fudge the $R\sin(\omega t)$ sign like I was since $-\omega t < 0$ in fourth quadrant?

Thanks!

 

[haruspex]

Ooooh thank you @haruspex!

I think I see what you are saying now. Am I please correct to say that we are not meant to fudge the $R\sin(\omega t)$ sign like I was since $-\omega t < 0$ in fourth quadrant?

Thanks!

Right.